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Subject: Statistical Question rss

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Brent Mair
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Roy
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I'm not the dullest tool in the shed when it comes to math but I never did learn much about statistics. Are there any websites that teach the basic stuff? I'm trying to figure out the statistical probably of a person drawing "Type W" card out of a deck of 66 when there are 16 "Type W" cards and the person draws 21 cards.

Thanks.
And when you're through with that, the train leaves Chicago heading west at 9:10 a.m. at 40 miles an hour...
 
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Ken Shogren
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Brent,

What you are looking for is an introduction to Probability (not statistics). Here's a link to an introductory book by a professor from Dartmouth College and Swarthmore College.

Introduction to Probability (free PDF book) -> http://www.dartmouth.edu/~chance/teaching_aids/books_article...

Good luck and enjoy.
-K
 
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Brent Mair
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Let me print out those 520 pages.

Thanks for the link. Maybe I'm dimmer than I thought.

I had meant to write statistical probability instead of statistics. Might still be wrong.
 
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Josh Luub
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Assuming you mean the chances of drawing at least one "type W" card...

This is one of those problems where it is easier to figure out the opposite. That is, what are the chances of drawing NO "type W" cards? Then you just subtract that answer from 1 and you get the answer you want.

So: If you draw one card, the chances of getting a non-type W card is 50/66 (there are 50 non-W cards in a 66 card deck). Assuming you drew a non-W card, then on your next draw you have a 49/65 chance of drawing a non-W card (there are only 49 non-W cards left, and only 65 cards left altogether).

so you multiply (50/66)*(49/65)*(48/64)*...*(29/45) (a total of 21 draws) to get the chance of drawing NO type-W cards. You get approximately .000756 (or 7 hundreths of 1%). Inverting that gives .99244, or in other words, about a 99.24% chance of drawing at least one type W card in 21 draws from a deck of 66 with 16 type W cards.

Alright, everyone got that? Now we'll do the same thing, but in base 8. Base 8 is just like base 10, if you're missing two fingers...



 
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Rod Spade
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One way to do it is with "combinations":

http://mathforum.org/dr.math/faq/faq.comb.perm.html
http://www.themathpage.com/aPreCalc/permutations-combination...

The formula for finding the number of combinations of k objects you can choose from a set of n objects is:

nCk = n! / [k! (n-k)!]

How many different ways are there to draw 21 cards from 66 (assuming you don't care about in which order you draw them)?

66C21

How many different ways are there to draw 21 of the 50 non-W cards?

50C21

So the probably of getting 21 non-W cards from the deck of 66 is:

50C21 / 66C21

If you write out all the factorials and simplify a bit, you'll see that it's exactly what squonk said. He iterated the draws one at a time, which is easier to understand. The "combinations" calculation lumps it all together, which may be quicker to calculate, but you'll probably have to simplify the expression to prevent your calculator from overflowing....


 
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